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Let $\displaystyle x=\sqrt{2}+\frac{b}{\sqrt{2}+\ldots}$, then \begin{align*} \sqrt{2}+\frac{b}{x} &= x \\ x\sqrt{2}+b &= x^{2} \\ x^{2}-\sqrt{2} \, x-b &= 0 \\ x &= \frac{\sqrt{2}+\sqrt{2+4b}}{2} \\ &= \sqrt{ \left( \frac{\sqrt{2}+\sqrt{2+4b}}{2} \right)^{2} } \\ &= \sqrt{b+1+\sqrt{2b+1}} \\ \end{align*} Take ...

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Just compute your left side: $$x=\sqrt{2}+\frac{b}{x}$$ $$x(x-\sqrt{2})=b$$ $$x^2-\sqrt2 x-b=0$$ $$x=\frac{\sqrt2+ \sqrt{2+4b}}{2}$$ So now you have $$\sqrt2+\sqrt{2+4b}=2\sqrt{a+c}$$ Can you take it from here?

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Let $$x = \sqrt 2 + \frac{b}{x}$$ Hence, $$x = \frac{\sqrt 2 x + b}{x} \implies x^2 = \sqrt 2 x + b \\ x^2 - x\sqrt2 + b = 0$$ The roots of this equation are $$x_1, x_2 = \sqrt2 \pm \frac{\sqrt{ \sqrt{2}^2 - 4b}}{2} = \sqrt{2} \pm \frac{\sqrt{2 -4b}}{2}$$ Let us consider the always-positive root $$x_1 = \sqrt{2} + \frac{\sqrt{2 -4b}}{2}$$ ...

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$x=(1;\overline{3,1})\implies x=1+\cfrac1{3+\cfrac1x}\implies3x^2-3x-1=0$. Thus, $$(1;\overline{3,1})=\frac{3+\sqrt{21}}6$$ Therefore, \begin{align} (2;\overline{1,3}) &=2+\frac6{3+\sqrt{21}}\\[3pt] &=\frac{1+\sqrt{21}}2 \end{align} $x=(3;\overline{3})\implies x=3+\cfrac1x\implies x^2-3x-1=0$. Thus, $$(3;\overline{3})=\frac{3+\sqrt{13}}2 ... 0 Quick answer: From link, you can use the formula$$[a,b] = \frac{-ab+\sqrt{ab(ab+4)}}{2a}.$$Then, you can check the result here with \frac{1+\sqrt{21}}{2} 3 Just an example: \sqrt{7}. Since 4<7<9, \left\lfloor \sqrt{7}\right\rfloor = 2, so:$$ \sqrt{7} = 2+(\sqrt{7}-2) = \color{blue}{2}+\frac{1}{\frac{\sqrt{7}+2}{3}}\tag{1}.$$Since 2+\sqrt{7}\in (4,5), \left\lfloor\frac{\sqrt{7}+2}{3}\right\rfloor =1, so:$$ \frac{\sqrt{7}+2}{3} = 1+\frac{\sqrt{7}-1}{3} = 1+\frac{1}{\frac{\sqrt{7}+1}{2}}$$and ... -1 Yes! In Mathematica, define Semiconvergent[a_, b_, k_: 1] := (Numerator[a] + k Numerator[b])/(Denominator[a] + k Denominator[b]) Then es = Convergents[E, 15] Semiconvergent[es[[13]], es[[14]], 5] Gives 271801/99990 1 I believe that for -\frac{39}{25} you should first rewrite it as -2+\frac{11}{25}. Then 11=0\times25+11 25=2\times11+3 11=3\times3+2 3=1\times2+1 2=2\times1+0 This gives -\frac{39}{25}=[-2+0;2,3,1,2]=[-2;2,3,1,2] Note that recently there has developed a way of representing negative continued fractions in the form ... 3 Let x be the value of the continued fraction [0;7]. We have that$$ x = \frac{1}{7 + \frac{1}{7 + \frac{1}{7 +...}}}$$Now we notice that x appears in this continued fraction, so we can write it as:$$x = \frac{1}{7 + x}$$This simplifies to x^2+7x-1=0, and solving for x gives x=\frac{-7}{2}\pm \frac{\sqrt{53}}{2} 1 Lehmer's procedure involves solving the Pell equation$$x^2-2qy^2=1$$for 173-smooth squarefree q\neq2. (This makes \frac{x}{y} an approximation for \sqrt{2q}.) Then, for a finite number of smallest solutions (x,y) of each Pell equation, the integers n=\frac{x-1}{2} and n+1 are tested for smoothness. Therefore, the x involved is x=2n+1, ... 4 Given your continued fraction F(x), it seems it has a general closed-form. Define,$$d=x^2+4\tag1$$then,$$F(x) = x-1-\frac{(x+1)\big(-x^3+12x+d\sqrt{d}\big)}{6(3x^2-4)}\tag2$$hence,$$G(x)=\frac{-x^3+12x+d\sqrt{d}}{6(3x^2-4)}=\cfrac{1}{x+\cfrac{(-1)(5)} {3x+\cfrac{(1)(7)}{5x+\cfrac{(3)(9)}{7x+\cfrac{(5)(11)}{9x+\ddots}}}}}\tag3$$Plugging in x, ... 1 Not a book exactly but free if you have access: Y.T. Cheng’s undergraduate thesis on continued fractions. 1 The scalar product (f,g) is defined in equation (2.2):$$ (f,g)=\int_{0^-}^{\infty} w(\lambda) \, f(\lambda) \, g(\lambda) \,d\lambda . $$Note that it includes the weight function w(\lambda). The polynomials p_0(\lambda), p_1(\lambda), p_2(\lambda),p_3(\lambda),\dots are what you obtain if you start with the monomials ... 1 I have had a glance at the paper. It is not that it is badly written but I consider pages 568-573 constitute a very intricated lecture on orthogonal polynomials, and the attached continued fraction. It is not surprising that you cannot easily find your way in this jungle. The presentation can be vastly simplified by centering all on the positive definite ... 2 Given the symmetric continued fraction found in this post$$\frac{\displaystyle\Gamma\left(\frac{a+3b}{4(a+b)}\right)\Gamma\left(\frac{3a+b}{4(a+b)}\right)}{\displaystyle\Gamma\left(\frac{3a+5b}{4(a+b)}\right)\Gamma\left(\frac{5a+3b}{4(a+b)}\right)}=\cfrac{4(a+b)}{a+b+\cfrac{(2a)(2b)} ...

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Generalized continued fractions can be either represented by Euler's continued fraction formula or Gauss's hypergeometric continued fraction which are both very general. I will give the OP one simple method which I often use (though I did not adopt it from any textbook but developed it independently) The idea is that every alternating series can be ...

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Short version: Continued fractions fall under Number theory. Long version: Continued fractions is one of the areas of mathematics that has applications in many other areas of mathematics, so it is difficult to place into just one branch. However, it has more easy applications to Number theory than to other branches of mathematics, so it is usually first ...

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I would start with Khinchin's Continued Fractions . For $8.95 you can't go wrong, and you can move on to something more advanced when you're done. http://store.doverpublications.com/0486696308.html 0 Look at what you've written (assuming the continued fraction converges): $$x=1+\cfrac1{\color{#C00000}{1+\cfrac1{1+\cfrac1{1+\dots}}}}\tag{1}$$ implies that $$x=1+\frac1{\color{#C00000}{x}}\tag{2}$$ which, in turn, implies that $$x=\frac{1+\sqrt5}2\quad\text{or}\quad x=\frac{1-\sqrt5}2\tag{3}$$ By transitivity,$(1)\implies(3)$. Since$x$, as defined ... 4 No, the negative number is not a solution. You showed that if$x$is equal to that fraction, then it is either$\frac{1+\sqrt 5}{2}$or$\frac{1-\sqrt5}{2}$. You calculated possible candidates for solutions, not the solution itself. You can prove that$x$must be positive by simply arguing that$x\$ is a limit of a sequence with only positive elements, so ...

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First some experimentation might be in order. I wrote up a program that could check for chains like this: program nr implicit none integer, parameter :: ik16 = selected_int_kind(38) integer, parameter :: N = 200 integer(ik16) p(N), q(N) integer D integer sqD integer r, s, a integer i, j, k integer(ik16) e, b, c ...

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